Explore topic-wise InterviewSolutions in .

Answer is Oxidation

Hydrogen resembles alkali metals in the combination with halogens ,oxygen and sulphur. On the other hand, hydrogen resembles halogens in that it exists as a diatomic molecule and it combines with metals and non-metals to form covalent compounds.

Molten KCl 

KCl (s) → K⁺ + Cl^- 

At the negative electrode: 

K⁺ + le^- → K (reduction – cathode) 

At the positive electrode: 

2Cl^- → Cl₂ + 2e^- (oxidation- anode)

Solution of potasium chloride. At the negative electrode:

2H₂O + 2e^- → H₂ + 2OH^- (cathode).

At the positive electrode:

2Cl^- → Cl₂ + 2e^- (anode).

a. Cu²⁺ , 

Cl^- , 

H₂O. 

b. 2H₂O → O₂ + 4H⁺ + 4e^- 

c. Na⁺ , 

Cl^- 

d. Na⁺ + le^- → Na 

e. 2Cl → Cl₂ + 2e^- 

f. 2H₂O + 2e^- → H₂ + 2OH^-

Towards negative electrodes(Cathode)

a. 

b. 

c. When Na⁺ ion and water are compared reduction occurs to water. Hence H is liberated at cathode.

H⁺ + e → H, 

H + H → H₂. 

d. Electrolyte. 

e. Electrolysis. 

f. Electroplating, Production of chemicals, Purification of metals.

1. Rate of forward reaction increases. 

2. Rate of forward reaction increases

3. Attains the equilibrium quickly. 

4. Rate of forward reaction increases

Gases are formed. CO₂ and SO₂ are the gases.

a. Since the number of ions is less, pure water does not allow the passage of electricity. 

b. When dil H₂SO₄ is added the water becomes a good electrolyte. Hence electricity passes through it. When a little dilute sulphuric acid is mixed with water large quantity of hydronium ions are formed.

c. Hydronium ion (H₃O⁺) 

d. H₂SO₄ → 2H⁺ + SO^2-₄ 

e. 2H⁺ + SO^2-₄  + 2H₂O → 2H₃O⁺ + SO^2-₄ 

f. Hydronium ion 

g. H₃O⁺ 

h. 2H₃O⁺ (aq) + 2e^- → H_2(g) + 2H₂O 

i. H₂O has the highest oxidation potential when compared to SO^2-₄ 

j. 2H₂O → O₂(g) + 4H⁺ + 4e^- 

k. 4H⁺ , SO₄^2- ions. 

l. H₂SO₄ is formed when these 2 combine together.

a. On increasing the pressure, yield of the product will be maximum. 

b. Increase the concentration of reactants nitrogen or hydrogen. Removal of product ammonia from the system.

The sulphuric acid removes water from the sugar in a highly exothermic reaction, releasing heat, steam, and sulfur oxide fumes. Aside from the sulfurous odor, the reaction smells a lot like caramel. The white sugar turns into a black carbonized tube that pushes itself out of the beaker. Dehydrating properties of H₂SO₄ are shown in the experiment. Concentrated sulphuric acid has the ability to absorb chemically combined water, or hydrogen and oxygen from substances in the ratio corresponding to that of water. This process is known as dehydration. Concentrated sulphuric acid is a strong dehydrating agent.

a. It’s character as a drying agent 

b. It’s character as a dehydrating agent.

Sulphuric acid is an important industrial chemical which is used in the manufacturing processes of many goods over a wide range of applications. Sulfuric acid used in pulp and paper industry for chlorine dioxide generation, tall oil splitting and pH adjustments. Sulfuric acid is a strongly acidic, oily liquid that may be clear to cloudy in appearance. Concentrated sulfuric acid acts as both an oxidizing and dehydrating agent.

Sulfuric acid was once known as oil of vitriol. Here are some of the growing number of end-users and applications using sulfuric acid is Agricultural chemicals Aluminum Sulfate, Batteries, Cellophane, Detergents, Explosives, Fertilizers, Gasoline, Herbicides, Iron and steel pickling, Jet Fuel, Kerosene, Leather, Lubricating Oils, Medicinal processes, Oil additives, Paper, Rayon and rubber, Sugar, Synthetic fibers, Veterinary drugs, Water softener regeneration, Water treatment, Yellow pigments.

a. Increase the concentration of oxygen, which increases the rate of forward reaction, 

b. When the concentration, pressure or tern perature of a system at equilibrium is changed, the system will readjust itself so as to nullify the effect of that change and attain a new state of equilibrium. This is Le Chatelier’s principle.

Correct option is D) used to

When Al-NaCl solution is electrolysed, H₂ is liberated at cathode, Cl₂ at anode and NaOH is formed in the solution. Hence pH of solution increases.

The alcohol which gives the most stable carbonium ion on dehydration is  (CH₃)₃C – OH

Phenol can be distinguished from ethanol by Sodium

The organic liquid ‘A’ is  C₂H₅OH

Starch is converted to ethanol by fermentation, the sequence of enzymes used is diastase, maltase, zymase 

Maltase, maltase, zymase  enzymes stepwise complete the fermentation reaction

2x² - 6x = 1

⇒ 2x² - 6x - 1 = 0

⇒ x = \(\frac{6\pm\sqrt{36+8}}4=\frac{6\pm2\sqrt{11}}4\) = \(\frac{3\pm \sqrt{11}}2\)

⇒ x² = \(\frac{(3\pm\sqrt{11})^2}4\) = \(\frac{9+11\pm6\sqrt{11}}4\)

 = 5 \(\pm\frac32\sqrt{11}\) 

 = \(\frac12\)(10 \(\pm\)3√11)

Case I:-

If x² = \(\frac12\)(10 + 3√11)

Then 1/4x² = 1/4 x \(\frac2{10+3\sqrt{11}}\) = \(\frac12\times\frac{10-3\sqrt{11}}{100-99}\) = \(\frac{10-3\sqrt{11}}2\)

∴ x² + 1/4x² = \(\frac12\)((10+3√11) + (10 - 3√11)) = 20/2 = 10

Case II:-

If x² = \(\frac12\)(10 - 3√11)

then 1/4x² = 1/4 x \(\frac2{10-3√11}\) = \(\frac12\times\frac{10+3√11}{100-99}\) = \(\frac{10+3\sqrt{11}}2\) 

∴ x² + \(\frac1{4x^2}=10\)

Hence, x² + \(\frac1{4x^2}\) = 10

A = \(\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}\)

\(\therefore\) A² = \(\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}\)\(\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}\)

 = \(\begin{bmatrix}5&-1&2\\9&-2&5\\0&-1&-2\end{bmatrix}\)

A² - 5A + 4I = \(\begin{bmatrix}5&-1&2\\9&-2&5\\0&-1&-2\end{bmatrix}\) - 5\(\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}\) + 4 \(\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\)

 = \(\begin{bmatrix}5-10+4&-1+0+0&2-5+0\\9-10+0&-2-5+4&5-15+0\\0-5+0&-1+5+0&-2+0+4\end{bmatrix}\)

 = \(\begin{bmatrix}-1&-1&-3\\-1&-3&-10\\-5&4&2\end{bmatrix}\)

Let A² - 5A + 4I + X = 0

Then X = -(A² - 5A + 4I)

 = \(-\begin{bmatrix}-1&-1&-3\\-1&-3&-10\\-5&4&2\end{bmatrix}\)

= \(\begin{bmatrix}1&1&3\\1&3&10\\5&-4&2\end{bmatrix}\)

The threshold frequency is defined as a minimum frequency under which the photoelectric emission is not possible.

the photoelectric effect is a phenomena where electrons emitted from the metal surface when the light of sufficient frequency is incident upon . wave theory says that light of any frequency should be capable of ejecting electrons.

Total letters = 11, I = 4, S = 4, P = 2. 

∴ The total number of permutations 7 

4s’s are together can be taken as 1 unit i.e.

∴ The number of permutations = 8!/(4! x 2!)

(b) 4s’s are not together = Total number of ways – 4s’s are together

= 11!/(4! x 4! x 2!) - 8!/(4! x 2!) (C)

Begin with MISS: The remaining 7 letters can be arranged in 7!/(3! x (2!)²)

(d) Begin with SIP: The remaining 8 letters (I = 3, S = 3) can be arranged in 8!/3!²

1. Since no two girls sit together, we have first arrange the 6 boys among themselves. This can be done in 6! ways.

Now no two girls sit together if we place the girls in between boys. There are 7 places and it should be occupied by 5 girls, can be done in 7P₅ ways. 

Therefore the total number of ways is 6! × 7P₅ 

= 720 × 7 × 6 × 5 × 4 × 3 

= 1814400.

2. Boys and girls occupy alternate position can be done as follows. 

First place the boys whose number is large.

Boys can be arranged in 6! ways. The place between boys can be filled by 5 girls, can be done in 5! ways. 

Therefore the total number of ways is 6! × 5! = 720 × 120 = 86400.

1. In the word MISSISSIPPI there are 11 letters, of which S appears 4 times, I appears 4 times, P appears 2 times and the rest all are different.

Therefore the total number of ways is \(\frac{11 !}{4 ! \times 4 ! \times 2 !}\)= 34650.

2. 4 I’s are kept together and should be counted as one unit, then there are 8 units. The number of ways is \(\frac{8 !}{4 \times 2 !}\) = 840. 

Therefore the I’s not come together 

= Total arrangements – 4 I’s together.

= 34650 – 840 = 33810.

1. DAUGHTER this word has 8 different letters. A, U, E are the vowels. Treat these 3 as one unit, then there are 6 units and can be permuted in 6! ways. The above vowels can be permuted in 3! ways. 

Hence the total number of words is 6! × 3! = 4320.

2. Number of words all vowels do not occur together = Total number of different words – number of words in which vowels come together

= 8! – 6! × 3! 

= 6!(8 × 7 – 6) 

= 2 × 6!(28 – 3)

= 50 × 6! = 36000.

Selecting 4 questions from Part I = 6C₄ = 6C₂

Selecting 4 questions from Part II = 7C₄ = 7 C₃

Selecting 4 questions from Part III = 8C₄

The required number of ways

= 6C₂ × 7C₃ × 8C₄ 

= 15 × 35 × 70 

= 36750.

et n be the member of person present in the meeting. The total number of handshakes is same as the number of ways of selecting 2 persons from among n persons.

The total number of handshakes = nC₁ = 45

⇒ \(\frac{n(n-1)}{1 \times 2}\) = 45 

⇒ n² – n – 90 = 0

⇒ (n – 10)(n + 9) = 0

⇒ n = 10, -9

n = 10 is acceptable.

n = 5 + 10 = 15

a⁵ = (1) (1/5)^{5 - 1} = (1/5)⁴ = [1/625]

Stage = m = - (3/-4) = [3/4]

A ∪ B = {3, 5} ∪ [2; 5, 6] = [2, 3, 5, 6] 

∴ (A ∪ B)’ = {1, 4}.

= n : a^n-1 = 3 (4)^3-1 = 3(4)² = 48

Sample space: Set of all possible outcome of an event is called a sample space.

“Every natural number is not greater than 0.” 

Answer is (A) sin(7π/18)+sin(4π/9)

\(\frac n2 + \frac n3 + \frac n5 = 31\)

⇒ \(\frac {15n + 10n + 6n}{30} = 31\)

⇒ \(\frac{31n}{30} = 31\)

⇒ \(n = 30\)

\(3(cos \frac \pi6 + i \,sin \frac \pi6) = 3e^{i\frac\pi6}\)   \((\because e^i\theta = cos \theta + i\, sin \theta)\)

or \(3(cos \frac \pi6 + i\, sin \frac \pi 6) = 3 (\frac{\sqrt3}2 + \frac i2)\)

\(= \frac 32 (\sqrt 3 + i)\)

coeff of x³ = 0

coeff of x² = 1

sum of coeff = 0+1 = 1 

The correct option is (C) 1